Consider the ring
\[\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}\]
and its ideal
\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}.\]
Show that $p$ is a prime ideal of the ring $\Z[\sqrt{10}]$.
An ideal $P$ of a ring $R$ is a prime ideal if whenever we have $ab \in P$ for elements $a, b \in R$, then either $a\in P$ or $b \in P$.
Proof.
Suppose that $a+b\sqrt{10}, c+d\sqrt{10} \in \Z[\sqrt{10}]$ and the product
\[(a+b\sqrt{10}) (c+d\sqrt{10}) \in P. \]
Then expanding the product, we have
\[ ac+10bd+(ad+bc)\sqrt{10} \in P.\]
Since $ac+10bd$ must be even number, we have either $a$ or $c$ is even.
Hence either
\[a+b\sqrt{10}\in P \text{ or } c+d\sqrt{10} \in P,\]
and we conclude that $P$ is a prime ideal.
Further Question.
In fact, it can be proved that $P$ is a maximal ideal in the ring $\Z[\sqrt{10}]$.
Prime Ideal is Irreducible in a Commutative Ring
Let $R$ be a commutative ring. An ideal $I$ of $R$ is said to be irreducible if it cannot be written as an intersection of two ideals of $R$ which are strictly larger than $I$.
Prove that if $\frakp$ is a prime ideal of the commutative ring $R$, then $\frakp$ is […]
Equivalent Conditions For a Prime Ideal in a Commutative Ring
Let $R$ be a commutative ring and let $P$ be an ideal of $R$. Prove that the following statements are equivalent:
(a) The ideal $P$ is a prime ideal.
(b) For any two ideals $I$ and $J$, if $IJ \subset P$ then we have either $I \subset P$ or $J \subset P$.
Proof. […]
Nilpotent Element a in a Ring and Unit Element $1-ab$
Let $R$ be a commutative ring with $1 \neq 0$.
An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.
Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.
We give two proofs.
Proof 1.
Since $a$ […]
If Every Proper Ideal of a Commutative Ring is a Prime Ideal, then It is a Field.
Let $R$ be a commutative ring with $1$.
Prove that if every proper ideal of $R$ is a prime ideal, then $R$ is a field.
Proof.
As the zero ideal $(0)$ of $R$ is a proper ideal, it is a prime ideal by assumption.
Hence $R=R/\{0\}$ is an integral […]
Characteristic of an Integral Domain is 0 or a Prime Number
Let $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.
Definition of the characteristic of a ring.
The characteristic of a commutative ring $R$ with $1$ is defined as […]
In a Principal Ideal Domain (PID), a Prime Ideal is a Maximal Ideal
Let $R$ be a principal ideal domain (PID) and let $P$ be a nonzero prime ideal in $R$.
Show that $P$ is a maximal ideal in $R$.
Definition
A commutative ring $R$ is a principal ideal domain (PID) if $R$ is a domain and any ideal $I$ is generated by a single element […]
$(x^3-y^2)$ is a Prime Ideal in the Ring $R[x, y]$, $R$ is an Integral Domain.
Let $R$ be an integral domain. Then prove that the ideal $(x^3-y^2)$ is a prime ideal in the ring $R[x, y]$.
Proof.
Consider the ring $R[t]$, where $t$ is a variable. Since $R$ is an integral domain, so is $R[t]$.
Define the function $\Psi:R[x,y] \to R[t]$ sending […]
[…] A direct proof that the ideal $P=(2, sqrt{10})$ is prime in the ring $Z[sqrt{10}]$ is given in the post “A prime ideal in the ring $Z[sqrt{10}]$“. […]
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[…] A direct proof that the ideal $P=(2, sqrt{10})$ is prime in the ring $Z[sqrt{10}]$ is given in the post “A prime ideal in the ring $Z[sqrt{10}]$“. […]